设 $a,b,c,d$ 为正实数，且 $a+b+c+d=4$．证明：\[\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\geqslant 4+(a-b)^2.\]
解析 齐次化，题中不等式即\[\sum_{\rm cyc}\left(\dfrac {a^2}b-a\right)\geqslant \dfrac{4(a-b)^2}{a+b+c+d},\]而\[LHS=\sum_{\rm cyc}\left(\dfrac{a^2}b+b-2a\right)=\sum_{\rm cyc}\dfrac{(a-b)^2}{b}\geqslant \dfrac{\left(\displaystyle\sum_{\rm cyc}|a-b|\right)^2}{\displaystyle\sum_{\rm cyc}b},\]而根据绝对值不等式，有\[|b-c|+|c-d|+|d-a|\geqslant |a-b|,\]于是\[\dfrac{\left(\displaystyle\sum_{\rm cyc}|a-b|\right)^2}{\displaystyle\sum_{\rm cyc}b}\geqslant \dfrac{\left(2|a-b|\right)^2}{\displaystyle\sum_{\rm cyc}b}=\dfrac{4(a-b)^2}{a+b+c+d},\]题中不等式得证．